Let $R$ be a commutative ring with unity. Let $C$, $D$ be complexes of $R$-modules, such that $C_m=D_m=0$ for all $m\lt0$.
Fixed $n\geq 0$, let $R(n)$ be the complex with $R(n)_n=R$ and $R(n)_m=0$ for $m\neq n$; let $R[n+1]$ be the complex with $R[n+1]_n=R[n+1]_{n+1}=R$ and $R(n)_m=0$ for $m\neq n,n+1$, with $\partial_{m+1}=1_R$. Any $t:R (n)\to R [n+1]$ will be assumed as the morphism with $t_n=1_R$.
Observation. An element of $C_{n+1}$ is the same as a morphism $R[n+1]\to C$; an element in $Z_n(C)$ is the same as a $R(n)\to C$. Plus a morphism $x:R(n)\to C$ factors in $y\circ t$, for a $y:R[n+1]\to C$, iff $x=\partial_{n+1}(y)$.
Let $p:C\to D$ be a morphism with the property ($*$) that, for all $n\geq 0$, given a commutative square $\require{AMScd}$ $$\begin{CD} R (n)@>>> C\\ @VVtV @VVpV \\ R [n+1]@>>> D \end{CD}$$ exists a morphism $R[n+1]\to C$ making the two resulting triangles commutative. I want to prove that $p$ is a weak equivalence showing that, for all $n$, given $x\in Z_n(C)$ and $y\in D_{n+1}$ such that $\partial_{n+1}(y) = p_n(x)$, exists $z\in C_{n+1}$ satisfying $\partial_{n+1}(z)=x$ and $ p_{n+1}(z) = y$.
So given $x\in Z_n(C)$, $y\in D_{n+1}$ such that $p_n(x)=\partial_{n+1}(y)$, the square below commutes. By ($*$) exists $z: R[n+1]\to C$ such that $p_{n+1}(z)=y$ (as $p\circ z=y$), and $\partial_{n+1}(z)=x$ (as $x=z\circ t$).
$$\begin{CD} R (n)@>x>> C\\ @VVtV @VVpV \\ R [n+1]@>y>> D \end{CD}$$
The problem is that I don't understand how to prove that the fact in italics implies that $p$ is a weak equivalence; it surely implies that $p_n^{-1}(B_n(D))\subset B_n(C)$ for all $n$, but shouldn't we prove also that all $Z_n(C)\to Z_n(D)$ are surjective to get a isomorphism on the homology modules? Thanks for any suggestion
Notice that $p : C\to D$ is a weak equivalence if and only if its cone is acyclic. To see this, put $\operatorname{cone}(p) = C \oplus D[-1]$ with differential $d(x,y) = (dx, px - dy)$, so that there is a short exact sequence of complexes
$$0 \to D[-1] \to \operatorname{cone}(p) \to C \to 0.$$
You can check that the connecting morphism $HC\to HD$ is $Hp$, so that $Hp$ is an isomorphism if and only if $H( \operatorname{cone}(p)) = 0$. This means that whenever $x\in C$ and $y\in D$ satisfy
$$\begin{align*}dy &= px \\ dx &= 0 \end{align*}$$
there exist $x'\in C$ and $y'\in D$ such that
$$\begin{align*} px' &= y + dy' \\ dx' &= x \end{align*}.$$
If you write down what datum the square you have gives you, and what datum the filler map gives you, and you should be able to see that the filling of that square gives what you want with $y'=0$, so a priori it is a stronger statement than $p$ being a weak equivalence.